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Physics48 (Futt 2011)
Chapter23: Gauss'Law
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"The only thing in life that is achievedwithout efort isfailure. " - Sourceunknown
"Weare what we repeatedlydo. Excellence,therefore,is not an act, but a habit." * Aristotle
"Acr as if whatyou do makesa dffirence, because
it does." - Sourceunknown
Reading: pages605- 620
Outline:
=
+
+
=
+
electric flux
Gauss'Law
Gauss'Law andCoulomb'sLaw
conductorsin electrostaticequilibrium
applicationsof Gauss'Law
cylindrical slmmetry
planar symmetry
sphericalsymmetry
ProblemSolvingTechniques
'
You shouldknow how to calculatethe electricflux tfuough a given surface.A useful fact to
rememberis: if the field is uniform, thenthe total flux througha closedsurfaceis zero.Sometimes
you can easilycalculatethe flux ofa uniform fieid throughpart ofa closedsurfacebut you have
beenaskedfor the flux throughthe ranaining part. Sincethe two contributionsto the total flux sum
to zero,they arethe negativesof eachother.
Someproblemsdeal with Gauss'law in the form ae(D= quno,
where(Dis the electricflux through a
q"no
closedsurfaceand
is the chargeenclosedby the surface.You might be giventhe chargeand
askedfor the flux or you might be given the flux (or the informationto calculateit) and askedfor the
enclosedcharge.
Whena problemdealswith a conductor,you shouldrememberthat the electricfield inside the
conductoris 0, that the field just outsidethe conductoris perpendicularto the surfaceand has
magnitudeo/e6,whereo is the areachargedensity,andthat the total chargeenclosedby any
Gaussiansurfacecompletelyinsidethe conductoris 0.
Manyproblemsaskyou to useGauss'lawtosolvefor the electricfield in givensituations.Become
adeptin finding an expressionfor the total flux througha Gaussiansurfacefor the threespecialcases
discussedin the text: cylindrical, spherical,andplanar symmetry.For a cylindrical Gaussiansurface,
^
t^
,.
with a radial field that is uniform over the curvedportion of the sr:rFace,I E . dA = (2trL)E , where r
surfacewith a
is the radiusofthe Gaussiancylinderand L is its length.For a sphericalGaussian
of the Gaussian
radiat field that is uniform over the swface,I n . d.l = (4nr2 )E , whercr is the radius
chargedalsity o
sphere.Also rememberthat the electricfield dueto a largeplanewith unifom area
is c/2to.
Mathenatical Skills
Electric Flux Calculations:
areaintegrals'If
To calculatethe electricflux, you shouldbe familiar with the evaluationof simple
to
i"t"gration is a rectanglewith sidesa and b, you will probablywant position the
tir","gi"r
the axes'The
.oordirrut "rsystemsothat the rcc1'n;le is in the a .yplanewith two of its sidesalong
flux integral
dy.
The
infinitesimalelementof areacanbelaken to be a rectanglewith sidesdx and
thenbecomes
drdy
E,(,.vr
I:=,1;=,
ofthe electric
Integrationsinx and] ale carriedout independently.If, for example,the z component
fieldis givenby E"=9t'Y+2Y.+hen
1b
/^?
* = Jloli=,pf lt + zy)a"ay= Jy+lzf f
I
tlo
+ ^2ry)l}=odv
= (o3y2
+ *\lr=r= o'b'+ ob',
= I!,=01u"'r'
+ ?ay)dy
y'
wherethe integtationoverx wascarriedout first, followed by the integrationover
For manyproblemsof this chaptera Gaussianswfacecanbe chosenso that the normal
on the
fi rft" electricfield hai the samevalueat all points on a portion of it andis zero
--p"*"t
=
region
of the
othe. portions.The integralfor the flux tlen reducesto @ EA, whercA is the area
the
normal
of
magnitude
the
E
is
and
ou", *hi"h th" ,rormalcomponentof the field is not zero
component,
The areaofa
To evaluatethe flux, you mustknow how to calculatethe areasofvarious surfaces'
zR2;the surfacearea
rectanglewith sidesof lengtha alldb is ab;the areaof a circle with radius-Ris
with
of a spiere with radiusR is 4rRz; andthes'rface meaof the rormdedportion of a cylinder
radiusR andlengthI is 2rRL.
imaginea
If you havetrouble rememberingthatLltL givestheareaof a cylindrical surface,
is the same
puprito*"i roU ','.uppedexactlyoice aroundwith a towel. The surfaceareaofthe roll
L' its areais
lritr" *"u of *t" towel. Sincethe towel is a rectanglewith sidesof lengSh'2r nd
2mL.
Calculationsof charge:
You must alsobe able to calculatethe chargeenclosedby a Gaussiansurfacewhen the volume,area"
or linear chmgedensityis given. If an object hasa uniform volume chargedensityp, for example,
the chargeenclosedis $venby pV, where Zis the volumeof the part of the object that lies within
the Gaussiansurface.Ifthe Gaussiansurfaceis completelywithin the object andthe object doesnot
haveany cavities,then the regionenclosedby the Gaussiansurfaceis completelyfi]led with charge
and lzis the volume enclosedby the Gaussiansurface.If the object hasa cavity that is wholly within
the Gaussiansurface,then Zis the volume enclosedby the Gaussiansurfaceminus the volume of the
ca\tity.
_ suppose,for example,a sphereof radiusR hasa uniform volume chargedensityp andthe
Gaussiansurfaceis a concentricsphereof radiusr.If r < R, then the Gaussiansphereis filled with
chargeandthechargeenclosedis 4npr3/3.If,on theotherhand..R< r, thenthe Gaussiansurfaceis
only partially filled with charge.The chargeenclosedis the total charge,or 4zpR3/3.
Ifa sphericalobject hasa sphericalcavity with radius.R"andthe Gaussiansurfaceis within the
objectbut outsidethe cavity, thenthe chargeenclosedis l(4r23 /3) - (42Rc3
B\. The first term in the
bracketsis the volume enclosedby the Gaussiansurfaceandthe secondis the volume of the cavitv.
IfR < r, the chargeenclosedis p[(4tn3B7 - 14rn"3/311.
Now, considera solid cylinderwith radiusR andlengthz, having a uniform volr,me ch.arge
densityp. Supposethe Gaussiansurfaceis a concentriccylinder with radiusr andthe samelengthas
the cylinder of charge.If r <.R,the chargeenclosedis 2zqL arfi,if R < r, itis 2npRL.If the rylinder
hasa cavity that is inside the Gaussiansurface,its volumemust be subtractedfrom the volumesin
theseexpressions,
If the volumechargedensityvariesfrom point to point in the object,you must evaluatethe
,f
rnte$al p dV over the volumeof that part of the object lying within the Gaussiansurface.The
J
most commonexampleis a spherewith a chargedensitythat variesonly with distancer from the
center.Carry out the integrationby dividing the sphereinto sphericalshellswith infinitesimal
thicknessdr. A typical shell extendsffom r to r + dr andhasa volume of 4zz2dr. Notice that this is
the productofthe surfaceareaofthe shell andits thickness.The chargein the shell is Axflr)rz fu.If
the Gaussiansurfaceis a sphereofradius r, concenhicwith the sphereof chargeand entirely within
- _ lr
it, the chargeenclosedisl 4z p(r) 12dr . If the Gaussiansurfaceis entirely outsidethe charge
distribution,tlre chargeenclosedis
lR
4tt p1r)r'1dr . Notice the upperlimits of integrationare
different for thesetwo cases.
For a cylinderwitl a chmgedensitythat dependsonly on the distancer from the axis, divide the
cylinder into concentriccylindrical shellswith thicknessdr. The volumeof a shell isZmL dr, wherc
Z is the lengthofthe cylinder.If the Gaussiansurfaceis a concentriccylinder with mdius r andis
insidetlre chargedistribution, thenthe chargeenclosedisJ" 2rrl p(r) dr. Ifthe Gaussiansurfaceis
outsidethe distribution,the chargeenclosedis
lR 2ttrL
p(r) dr .
Questions and Example Problems from Chapter 23
.ball lieswithinthehollowof a metallicsphericalball of radiusR. Forthreesituations,
fliililJ*r"o
are(1)+4q,0;Q) -4q, +10q; (3)+16q, -12q.Rank
ontheballandshell,respectiveTy,
thenetcharges
to thechargeon (a)theinnersurfaceof theshelland(b) theoutersurface,most
thesituationsaccording
oositivefirst.
( r)
Z,^\
(3)
(A)
(@))
\/
f,^,fn-q -aArAQ
- 4 o' D
,c/dn ,\A-sIl
o,Jb
-t fu-i-t '" ' (t%
,,Ln"ur.'q
' +l{0.6
d4&4Are-t
-,rrrur-o,z.D.q!t :
n',fy IWU'.
- (6t
+ r.lq)
(r)
(al
Problem I
The squaresurfaceshownin the figure below measures3.2mm on eachside.It is immersedin a
rmiform electricfie1dwith magnitudeE = 1800N/C. The field lines makean angleof 35owith a
normal to the surface,as shown.Takethat normal to be directed"outward," asthoughthe surface
were one faceofa box. Calculatethe electricflux throughthe surface.
=>
*n
o 4f:,J' -:,,.l.r*D.Ke-e$ o''\je A b\
-e
e .Lr.rrirpL,ni -e-0-p-clt
{vsAJ
-.t ip r,tsfi^rit-
u-"f,-,
'l*;.
Do N/,^
F = )f
'
te
fi = (3.2.rlot.)'
,rr"
-t
&fr.rx
)"t
o$ n*'a'-g*'-ti U Et
,1ur-e^11)rA'"
-)
Q= E.A'= eAr-oO
j,4
f,,^€,.{
F * e.'
x @: J1,.*"?tu-r'd
; e= 115"t@25'
/
O= EAcaeO
=
6=
( trooN/c) (2"a" lot n,)\ co':115"
-/.c *1o'rln'/c,
Problen 2
cubeof length2.0 m and
An electricfield givenby E = 4.0i -3.O(y'z+2.Qj piercesa Gaussian
positionedasshownin the figure below. (The magnitudeof E is in N/C andposition x is in m.) What
is the electricflux t}rough the (a) top face,(b) bottom face,(c) left face,and (d) back face?(e) What
is the net electric flux throughthe cube?
4.ot - 3.o (ya-+a.o)i
E
t
v
afr0.l9
aa
",J,AaAA It- ,ttYi}J;ot
p=
( q.o rs/") i - (3.oN/c)
.t-
(")
ly"tleip W
I y'cmt* e"o]5.
--, ) - A.or.,
E= 4,Dt - 3.ofCr.";\ l"ol1
= 4,oi - tr1
'--t
-->
0=) a"tn'
/
( ot A-= (aaN/,-)C-(ren,l.)j)
. -'t
JA',=
JAj *^bPry"^
- Grr)rn = -lrN/c5{A
5(r.02-re3)-(J*1) f
5dn
O=
= 4"o rnr-\ot- c",oa t4
4
W
-/rN/c (+.on'):[ % N'.'l
\/ - /\ra^),il"r"1 t -llt fu4
Jn(-e) i..i=o
Q= \ E.tr = \ lU"ot- (;'o)(vla"")t]"
( ( q,ot\ ,Jnte)
),,/
-'-
-16Nm'./^a
CA=
Y
/t'
= -L/.oft^ = -4"ory.(4"D
")
Problem3
A chargedparticle is suspendedat the cent€rof two concentricsphericalshellsthat arevery thin and
madeofnonconductingmaterial.Figure a below showsa crosssection.Figureb givesthe net flux
O througha Gaussianspherecenteredon the particle, asa function of radiusr of the sphere.The
scaleof the vertical axis is set by @s=5.0 x 105Nmz/C.(A) What is the chargeof the central
particle?What arethe net chargesof @) shell A and (c) shell B?
_o.
fue* )4r., ^'
:
'Y
,R-
?0
DF
j-d.u))
= nn*tL,
"dn
-$.n.) Q a
--C'
rA
h
a) -{-orr< ro
,; +'n
l,oxp5v1^x7,
..n).gr"n.t
, Q=
W
FC
-lD
=
(!.oroqn,,'fd (t"zsrtoloc/N,r,')ll'??^
I
$nn"l 0u
b) $^ e,\< c. h ., d: -5.o"ro5N*7. ,r"*-A1r",*t- r,t'.t
-i C6
A
= -4"q3*lo-*L
.
$*rr.) (-s"oxlD5N rl/c)(t.zs"lortrc'/p*';
IUur--*'-1_.:_1--.-.._.-------T_"1
q''l'i^lo-"Cl
Lc) -oyr-rIalq r l% *=
probtem
4
><I0+eG-side-fu conductoris a
An isolatedconductorof arbitiry .hup"h* u o"t""lhar!e'ofTT0
cavitywithin whichis a pointchargeq : +3.0x 10' C. Whatiis thecharge(a) onthqcavitywall and
(b) on the outer surfaceofthe conductor?
(")
-.U^t^uAt-'"
tu$'-t*
cAaoa-c,n Pt*'t''aa o'"ui'+a/.€
qJ*
i*.,, t.*''J-'r'*'fn
-,t;)s"*
-S;"r"" E=o
".*}xd "
Yu dn4)"r-c $,-lJ
}}$
+
L&crAA r*^a!
rcr^J-LLr!'Ua
).A d?il)a*4
)3Y -2,lr",&^'u
( b)
^'t q
5-r.
"atL$an
^
<cvnAur#t,
^1 ffw h-'n--
)r-tJt
^ ryuo ^. qE,J A'= O. s)e4rYv
. -.\
b;:Y]:'.
o\L
n
- /-r ,
Llen") = (++
O '*.l.F.J
-.. r-. -= Q
b
wd.q
fi,*
^Aol'tr'uz'an^t"L
c-!,.oo6, Q on l3' rc,o"d,u-*,
i
%**& = -X
t^)a{P
A u = | 3 " Or ' r . '
Problems
/
Thefigure.belowis a sectionofa conductingrod of radiusRr : 1.3QlmmandlengthL: i 1.00m
insidea thin-walled coaxial conductingcylindrical shell of radiusRz: 10.0Rrandthe (same)length
L. Thenet chargeon the rod is Q, : +3.40x 10rz C; that on the shellis Qz = -2.00Qr. Whatarethe
(a) magnitudeE and(b) direction (radially inward or outward)ofthe electric field at radial distance
r = 2.00R2?Whatare(c) E and(d) the directionat r: 5.00Rr?Whatis the chmgeon the (e) interior
and(f) exterior surfaceofthe shell?
f F. lA'= t
.
-iyrao .oad+
E(),nrL)
E
(a)
!>ot
,-- \.ooA"-
I
] . t 1 o * 1 o ' ' c - + ( - { o . f o ^l o ' ' c )
-4,1'u^^ Io-'tc-
,,;G;;''.%^.) ( a a , r d 3 ^ ) (l l - o " . )
(b)
C =5-ooAr
9. eo'r
0.u €o
(+)
d\tt Zo
- la^
I
/e\
=
QrtC{2,
-'
--
3.{o rlO L
F= O.Irr g7
/ t/
4:t
/\
'=
rL
-- I
d' tt (o
fc)
E(a'.cL)
- q. *'/t-
d\ lr go
=
5E.Jf,=
*gr'" I
-/
cb"n"l=
)
.".,4.
l
S.."r
=
$.n.\= O r 3''iO* IO-'"cI
3.,tol lC'aC
-rE
rL
Q-.*
\
Nir( 7.f 5'lo /1m')
= -Q, =
* po'^I ( rr'o".)
( r..so
- 3.riO ^
lD-'ac
--J..1o"lo-'au
<o )
Problem6
A squarerretal plate of edgelength 8.0 cm andnegligiblethiclinesshasa total chargeof 6.0 x10'6C.
(a) Estimatethe magnitudeE ofthe electricfield just offthe centerof the plate (al say,a distanceof
0.50mm) by assumingthat the chargeis spreadunifomrly over the two facesof the plate.
h,t
E = V€.ulq-p"a
=) -a,vnxr ;)t}o J'^^4t
/-
=
rfl
w/r
t
E-
=
Co
A = -4^r4\a4! uJ'^nZu ds"r"'*'d 2f V* t:"*$*'q.
,^, -a4r'rl.aA.,u^"+zr*'l,6 o"a#v
i9,, a*v$^n
"o.-.qS "d
(-
+ <'e$u""b"'y 14r,/+a/'4tz 'ttr,*
4n'* ^A*
(o) J)-rld;rt+^'!d
-ao , = (ry
.7t trt }i*a! g's"a^€'t
(ry=
;
*t"o-p'.a-u , ;trv
n
--))"
AA
=
*
(R= ^')
=
F.g^,"",-/
6.o^ ld'c
A (g.o^to-a-)t(t.vs^tf'"c"A^')
ili#"U?l" *tow, a smallcircularholeof radiusR = l 80crnhasbeencutintotlremicldleof an
infinite flat, nonconductingsurfacetlpt basuniform chargedensityo : 4.50 pC/m". A z axis, with
its origin at the hole's center,is perpendicularto the sutface.In unit vectornotation, what is the
electricfield at point P at z : 2.56 cm?
^);h a /wAr-. A "1r.f ,J,a"4'
>
ri"rtu-*rAub*A4,0.De+
f -- 7aU
) E=9au(t:*6
L-fMT
-a
E^r-**t, &)
,.
e_+
. A€-I zTa,
(R)
( q.so^ | d''c/.*) 1 o.o15/o'"')
A ( a.ts * I d'' c/*'") { {o.oasr"-)\1o,orr'.)"
F:(o"^"€N/.)R
>al-.aL.Xl*
tPtu"*
Problem 8
Two chargedconcentricsphereshaveradii of 10.0cm and 15.0cm. The chargeon the inner sphere
is4.00xlOaC,andthatontheoutersphereis2.00xl0-sC.Findtheelectricfield(a)atr:12.0cm
and(b) at r : 20.0cm.
().-
-+
= a.rl
ln
nn"Ye.
E"
\
f4. ln=%u-tk_
E Ulxr")= s..-/€"----) E =akL 3Tf
.\
(o\)
= lA,ocro,
=4.oorld8C
$es.)
4^.
( r . 9 q x ) o q N r l 7 c a ) C e r " o olxo j e )
A-
(t)" ier.,)a
( u) -Vn r = lo.o*o,
?
6c -- 6"oo,. lo ec
=
$ encl { . o o x 1 9 - t g + 3 . . o o x l o
= 1t"=s*io+N,/rI
(o.aorn)\
!'-t
Problem 9
A solid nonconductingsphereofradius R = 5.60 cm hasa nonuniformchargedistributionofvolume
chmgedensityp: (14.1 pclmi) r/R, wherer is the radial distancefrom the sphere'scenter.(a) What
is the sphere'stotal charge?What is the magrritudeE of the electricfield at (b) r: 0, (c) r = R/2.00,
and(d) r: R?
.
\--'l
71I'-lf-/rn
p = Ar/o
/
\
c! =
(")
\
tl
A=fi.lxlo
Q(c)Jv
%= Ll4rAR
_:---
=
(+)
'1.7t" ld'=c
(.)
t/*'
A-
S.Lo.r'.,
-6
A
= tJA('coJ"
'"
'''
t'(A7n)4nr'dc
o
h
= rrAB' : -(l,l.r^10'"c/.)(t.(,o^tdl-,)3
-" --lFl
( b) .t p=o,(eo")
f F.rtr=
.-. E
A (rync)= (u*)t%,)
4""ye. _-+ f (anc) '= \*U,>"BLee*r= 1"j
r-- -- ---!-
(d)
*-. f=A , S,*r= ?.?9"10'sc
4oe-
-rA r'
B
c
( r"d"